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A breeder of show dogs is interested in the number of female puppies in a litter. If a birth is equally likely to result in a male or a female puppy, give the probability distribution of the variable \(x=\) number of female puppies in a litter of size 5 .

Short Answer

Expert verified
The probability distribution of the variable \(x=\) number of female puppies in a litter of size 5 is: \(P(0) = 0.03125\), \(P(1) = 0.15625\), \(P(2) = 0.3125\), \(P(3) = 0.3125\), \(P(4) = 0.15625\), \(P(5) = 0.03125\).

Step by step solution

01

Defining the Variables

The variable \(x\) is defined as the number of 'successes', which in this case is the number of female puppies. Since a litter includes 5 puppies, \(x\) can be any integer from 0 to 5. The probability of 'success' (in this case, having a female puppy) is 0.5, since it's mentioned that a birth is equally likely to result in a male or a female puppy.
02

Applying the Binomial Formula

The probability \(P(x)\) of having \(x\) female puppies in a litter of 5, where the probability of each puppy being female is 0.5, is given by the binomial formula: \[P(x) = C(n, x) \cdot (p)^x \cdot (1-p)^{n-x}\]. In this case, \(n\) is 5 (size of the litter), \(x\) is the variable we are calculating the probability for, \(p\) is 0.5 (probability of a puppy being female), and \(q = 1-p = 0.5\) is the probability of a puppy being male. \(C(n, x)\) is the number of combinations of \(n\) items taken \(x\) at a time, and can be calculated using the formula \[C(n, x) = \frac{n!}{x!(n-x)!}\].
03

Calculating the Probabilities

We calculate \(P(x)\) for each possible value of \(x\), from 0 to 5, using the binomial formula. The results are: \(P(0) = 0.03125\), \(P(1) = 0.15625\), \(P(2) = 0.3125\), \(P(3) = 0.3125\), \(P(4) = 0.15625\), \(P(5) = 0.03125\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions. Specifically, it deals with the number of successes in a fixed number of trials, given the same probability of success in each trial.

For example, in the case of the breeder's interest in the number of female puppies per litter, the distribution can tell us the probability of getting a certain number of female puppies out of five births, assuming each birth is as likely to be male as it is female.

An important feature of the binomial distribution is its predictability. Based on the fixed probability of a single event occurring, it can be used to determine the likelihood of various outcomes over a larger number of events, making it extremely useful in fields such as genetics, quality control, and survey analysis.
Combinatorics
The field of combinatorics is central to understanding binomial distributions, as it involves the counting of combinations and permutations. It helps calculate the number of different ways we can see a given number of successes (or specific outcomes) from a series of events.

Counting Combinations

In our puppy example, combinatorics answers the question: 'In how many different ways can we have a certain number of female puppies out of five?'. The formula for calculating combinations is given by \[C(n, x) = \frac{n!}{x!(n-x)!}\], where '!' denotes a factorial, the product of all positive integers up to that number.

This becomes pivotal when determining the different possible scenarios that can occur within our trials, which in turn is a key step in calculating the probabilities using the binomial formula.
Probability of Success
In a binomial distribution, the 'probability of success' is a fundamental component. It is the chance that a given trial (or event) will result in the outcome defined as 'success'.

In our dog breeding scenario, the 'success' is defined as the birth of a female puppy. Since the problem states that a birth is equally likely to result in a male or female puppy, the probability of success (\(p\)) is 0.5. Likewise, the probability of 'failure' (the birth of a male puppy in this case) is also 0.5, because the two potential outcomes are mutually exclusive and exhaustive.

Understanding the probability of success is crucial since this value is plugged into the binomial formula to calculate the probability of various outcomes across trials.
Binomial Probability Formula
The binomial probability formula is a concise mathematical expression that combines principles of combinatorics and the probability of success to yield the probability of achieving exactly \(x\) successes in \(n\) independent trials. It's presented as: \[P(x) = C(n, x) \cdot (p)^x \cdot (1-p)^{n-x}\].

Here, \(C(n,x)\) calculates the number of ways \(x\) successes can occur out of \(n\) trials, \(p\) is the probability of success for any individual trial, and \((1-p)\) is the probability of a single trial resulting in 'failure'. Each term in this formula has a specific purpose:
  • \(C(n, x)\) accounts for the combinatorial aspect; the different combinations of outcomes.
  • \(p^x\) gives the likelihood of achieving success \(x\) times.
  • \((1-p)^{n-x}\) accounts for the instance of 'failure' occurring in the remaining \(n-x\) trials.
By calibrating these values for the situation of the number of female puppies in a litter of five, one can predict the full range of possible outcomes with their corresponding probabilities.

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Most popular questions from this chapter

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. (Hint: See Example 6.32 ) a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109\).) d. \(P(75 \leq x \leq 125)\)

A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to customers in 5 -pound lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lcccc} x & 1 & 2 & 3 & 4 \\ p(x) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of \(x\).

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